Question

Eliminating arbitrary constants $a$ and $b$, find the differential equation represented by the family of curves: \[ y^2 = a(b^2 - x^2) \]

Hint:

For families of curves with two arbitrary constants, differentiate twice to eliminate them and obtain the required differential equation.

Answer 1

Step 1: Given family of curves. \[ y^2 = ab^2 - ax^2 \]

Step 2: Differentiate w.r.t. $x$. \[ 2y \frac{dy}{dx} = -2ax \] \[ a = -\frac{y}{x}\frac{dy}{dx} \]

Step 3: Substitute $a$ in original equation. \[ y^2 = \left(-\frac{y}{x}\frac{dy}{dx}\right)(b^2 - x^2) \]

Step 4: Eliminate $b^2$. From the given: \[ y^2 + ax^2 = ab^2 \] Differentiate w.r.t. $x$: \[ 2y \frac{dy}{dx} + 2ax = 0 \] Which we already had. To eliminate $b$, differentiate again.

Step 5: Differentiate again. \[ 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2} + 2a = 0 \] Now eliminate $a$ using $a = -\frac{y}{x}\frac{dy}{dx}$. \[ 2\left(\frac{dy}{dx}\right)^2 + 2y\frac{d^2y}{dx^2} - 2\frac{y}{x}\frac{dy}{dx} = 0 \] Divide by 2: \[ \left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} - \frac{y}{x}\frac{dy}{dx} = 0 \]

Final Answer: \[ \boxed{\left(\frac{dy}{dx}\right)^2 + y\frac{d^2y}{dx^2} - \frac{y}{x}\frac{dy}{dx} = 0} \]